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4 years ago

The class is 60% women and 10% of the women have blond hair, so blond women comprise 60% × 10% = 6% of the class.

Mathematics

1 answer:

Tatiana [17]4 years ago

6 0

Answer:

Conversion and substitution

Step-by-step explanation:

One way to find out if the expression is true, we can convert all the percentages into their decimal equivalence.

60% = 0.6

10% = 0.1

6% = 0.06

Now let's substitute the values.

0.06 = 0.6 x 0.1

0.06 = 0.06

Since 0.06 is equal 0.06, this then proves that it is true.

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sergeinik [125]

Answer:

sample mean:: 17.92

ME = z*s/sqrt(n) = 1.8807*6.35/sqrt(47) = 1.7421

------------------------

Step-by-step explanation:

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3 years ago

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3 years ago

En una tienda de electrodomesticos , una nevera con un precio de 260€ fue rebajada un 15% ; al cabo de un mes volvieron a rebaja

Mamont248 [21]

Answer:

Part 1) The price after the second discount is 187.85€

Part 2) The percentage of total variation is 27.75%

Step-by-step explanation:

The question in English is

In an appliance store, a refrigerator with a price of 260€ was discounted by 15%; after one month it was discounted by another 15%. What is the price after the second discount? What is the percentage of total variation?

Part 1)

Find out the price after the second discount

we know that

$15\%=15/100=0.15$

A) The first discount is equal to

260€(0.15)=39€

The price after the first discount is equal to

260€-39€=221€

B) The second discount is equal to

221€(0.15)=33.15€

The price after the second discount is equal to

221€-33.15€=187.85€

Part 2) Find out the percentage of total variation

we know that

The original price of 260€ represent the 100%

by using proportion

Find out the percentage that represent (260€-187.85€)=72.15€

$\frac{260}{100}=\frac{72.15}{x}\\ \\x=72.15*100/260\\ \\x=27.75\%$

8 0

4 years ago

At Store A, a 60-ounce bag of shelled walnuts costs $11.99. At Store B, a 4-pound bag of shelled walnuts costs $11.99. Shelled w

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5 0

3 years ago

Read 2 more answers

After a college football team once again lost a game to their archrival, the alumni association conducted a survey to see if alu

valentina_108 [34]

Answer:

P-value for this hypothesis test is 0.00175.

Step-by-step explanation:

We are given that the alumni association conducted a survey to see if alumni were in favor of firing the coach.

A simple random sample of 100 alumni from the population of all living alumni was taken. Sixty-four of the alumni in the sample were in favor of firing the coach.

Let p = proportion of all living alumni who favored firing the coach

SO, Null Hypothesis, $H_0$ : p = 0.50 {means that the majority of alumni are not in favor of firing the coach}

Alternate Hypothesis, $H_A$ : p > 0.50 {means that the majority of alumni are in favor of firing the coach}

The test statistics that will be used here is One-sample z proportion statistics;

T.S. = $\frac{\hat p-p}{{\sqrt{\frac{\hat p(1-\hat p)}{n} } } } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of the alumni in the sample who were in favor of firing the coach = $\frac{64}{100}$ = 0.64

n = sample of alumni = 100

So, test statistics = $\frac{0.64-0.50}{{\sqrt{\frac{0.64(1-0.64)}{100} } } } }$

= 2.92

Now, P-value of the hypothesis test is given by ;

P-value = P(Z > 2.92) = 1 - P(Z $\leq$ 2.92)

= 1 - 0.99825 = 0.00175

Therefore, the P-value for this hypothesis test is 0.00175.

4 0

4 years ago